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Disclaimer: After a bit of work, I've come up with the following example. Happy to hear about others.

Fix $\alpha \in (1,\infty)$, and for any integer $n \ge 1$, let $\lambda_n = n^{-\beta}$, and let $\Lambda = \Lambda(n)$ be the $n \times n$ diagonal matrix with diagonal entries $\lambda_1,\ldots,\lambda_n$. We refer to this model for $\Lambda$ as $\beta$-polynomial.

Thanks to Example 4 of this paper From Gauss to Kolmogorov: Localized Measures of Complexity for Ellipses, we know that $w(S) \asymp R^{1-1/\beta}$. On the other hand, one computes $\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)}) \asymp \min(R\sqrt{n},1)$. We deduce that,

If $n^{-\beta/2} \ll R \lesssim n^{-1/2}$, then $$ \frac{\omega(S)}{\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)})} \asymp \frac{R^{1-1/\beta}}{R\sqrt{n}} = \frac{1}{R^{1/\beta}\sqrt n} \ll 1. $$

Disclaimer: After a bit of work, I've come up with the following example. Happy to hear about others.

Fix $\beta \in (1,\infty)$, and for any integer $n \ge 1$, let $\lambda_n = n^{-\beta}$, and let $\Lambda = \Lambda(n)$ be the $n \times n$ diagonal matrix with diagonal entries $\lambda_1,\ldots,\lambda_n$. We refer to this model for $\Lambda$ as $\beta$-polynomial.

Thanks to Example 4 of this paper From Gauss to Kolmogorov: Localized Measures of Complexity for Ellipses, we know that $w(S) \asymp R^{1-1/\beta}$. On the other hand, one computes $\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)}) \asymp \min(R\sqrt{n},1)$. We deduce that,

If $n^{-\beta/2} \ll R \lesssim n^{-1/2}$, then $$ \frac{\omega(S)}{\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)})} \asymp \frac{R^{1-1/\beta}}{R\sqrt{n}} = \frac{1}{R^{1/\beta}\sqrt n} \ll 1. $$

Disclaimer: After a bit of work, I've come up with the following example. Happy to hear about others.

Fix $\alpha \in (1,\infty)$, and for any integer $n \ge 1$, let $\lambda_n = n^{-\beta}$, and let $\Lambda = \Lambda(n)$ be the $n \times n$ diagonal matrix with diagonal entries $\lambda_1/s_n,\ldots,\lambda_n/s_n$, with $s_n := \sum_{1 \le k \le n}\lambda_k \to \zeta(\beta) := \sum_{k \ge 1}k^{-\beta} < 1$. By construction, it is clear that $\mbox{trace}(\Lambda)=1$. We refer to this model for $\Lambda$ as $\beta$-polynomial.

Thanks to Example 4 of this paper From Gauss to Kolmogorov: Localized Measures of Complexity for Ellipses, we know that $w(S) \asymp R^{1-1/\beta}$. On the other hand, one computes $\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)}) \asymp \min(R\sqrt{n},1)$. We deduce that,

If $n^{-\beta/2} \ll R \lesssim n^{-1/2}$, then $$ \frac{\omega(S)}{\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)})} \asymp \frac{R^{1-1/\beta}}{R\sqrt{n}} = \frac{1}{R^{1/\beta}\sqrt n} \ll 1. $$

Disclaimer: After a bit of work, I've come up with the following example. Happy to hear about others.

Fix $\alpha \in (1,\infty)$, and for any integer $n \ge 1$, let $\lambda_n = n^{-\beta}$, and let $\Lambda = \Lambda(n)$ be the $n \times n$ diagonal matrix with diagonal entries $\lambda_1,\ldots,\lambda_n$. We refer to this model for $\Lambda$ as $\beta$-polynomial.

Thanks to Example 4 of this paper From Gauss to Kolmogorov: Localized Measures of Complexity for Ellipses, we know that $w(S) \asymp R^{1-1/\beta}$. On the other hand, one computes $\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)}) \asymp \min(R\sqrt{n},1)$. We deduce that,

If $n^{-\beta/2} \ll R \lesssim n^{-1/2}$, then $$ \frac{\omega(S)}{\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)})} \asymp \frac{R^{1-1/\beta}}{R\sqrt{n}} = \frac{1}{R^{1/\beta}\sqrt n} \ll 1. $$

Disclaimer: After a bit of work, I've come up with the following example. Happy to hear about others.

Fix $\alpha \in (1,\infty)$, and for any integer $n \ge 1$, let $\lambda_n = n^{-\beta}$, and let $\Lambda = \Lambda(n)$ be the $n \times n$ diagonal matrix with diagonal entries $\lambda_1/s_n,\ldots,\lambda_n/s_n$, with $s_n := \sum_{1 \le k \le n}\lambda_k \to \zeta(\beta) := \sum_{k \ge 1}k^{-\beta} < 1$. By construction, it is clear that $\mbox{trace}(\Lambda)=1$.

Thanks to Example 4 of this paper From Gauss to Kolmogorov: Localized Measures of Complexity for Ellipses, we know that $w(S) \asymp R^{1-1/\beta}$. On the other hand, one computes $\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)}) \asymp \min(R\sqrt{n},1)$. We deduce that,

If $n^{-\beta/2} \ll R \lesssim n^{-1/2}$, then $$ \frac{\omega(S)}{\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)})} \asymp \frac{R^{1-1/\beta}}{R\sqrt{n}} = \frac{1}{R^{1/\beta}\sqrt n} \ll 1. $$

Disclaimer: After a bit of work, I've come up with the following example. Happy to hear about others.

Fix $\alpha \in (1,\infty)$, and for any integer $n \ge 1$, let $\lambda_n = n^{-\beta}$, and let $\Lambda = \Lambda(n)$ be the $n \times n$ diagonal matrix with diagonal entries $\lambda_1/s_n,\ldots,\lambda_n/s_n$, with $s_n := \sum_{1 \le k \le n}\lambda_k \to \zeta(\beta) := \sum_{k \ge 1}k^{-\beta} < 1$. By construction, it is clear that $\mbox{trace}(\Lambda)=1$. We refer to this model for $\Lambda$ as $\beta$-polynomial.

Thanks to Example 4 of this paper From Gauss to Kolmogorov: Localized Measures of Complexity for Ellipses, we know that $w(S) \asymp R^{1-1/\beta}$. On the other hand, one computes $\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)}) \asymp \min(R\sqrt{n},1)$. We deduce that,

If $n^{-\beta/2} \ll R \lesssim n^{-1/2}$, then $$ \frac{\omega(S)}{\min(R\sqrt{n},\sqrt{\mbox{trace}(\Lambda)})} \asymp \frac{R^{1-1/\beta}}{R\sqrt{n}} = \frac{1}{R^{1/\beta}\sqrt n} \ll 1. $$